Answer:
a. 0.4544 N
b. [tex]5.112 \times 10^{-5 M}[/tex]
Explanation:
For computing the normality and molarity of the acid solution first we need to do the following calculations
The balanced reaction
[tex]H_2SO_4 + 2NaOH = Na_2SO_4 + 2H_2O[/tex]
[tex]NaOH\ Mass = Normality \times equivalent\ weight \times\ volume[/tex]
[tex]= 0.3200 \times 40 g \times 21.30 mL \times 1L/1000mL[/tex]
= 0.27264 g
[tex]NaOH\ mass = \frac{mass}{molecular\ weight}[/tex]
[tex]= \frac{0.27264\ g}{40g/mol}[/tex]
= 0.006816 mol
Now
Moles of [tex]H_2SO_4[/tex] needed is
[tex]= \frac{0.006816}{2}[/tex]
= 0.003408 mol
[tex]Mass\ of\ H_2SO_4 = moles \times molecular\ weight[/tex]
[tex]= 0.003408\ mol \times 98g/mol[/tex]
= 0.333984 g
Now based on the above calculation
a. Normality of acid is
[tex]= \frac{acid\ mass}{equivalent\ weight \times volume}[/tex]
[tex]= \frac{0.333984 g}{49 \times 0.015}[/tex]
= 0.4544 N
b. And, the acid solution molarity is
[tex]= \frac{moles}{Volume}[/tex]
[tex]= \frac{0.003408 mol}{15\ mL \times 1L/1000\ mL}[/tex]
= 0.00005112
=[tex]5.112 \times 10^{-5 M}[/tex]
We simply applied the above formulas
The volume of the 0.3200 M, NaOH required to neutralize the H₂SO₄, is
21.30 mL, which gives the following acid solution approximate values;
1) Normality of the acid solution is 0.4544 N
2) The molarity of the acid is 0.2272
How can the normality, molarity of the solution be found?Molarity of the NaOH = 0.3200 M
Volume of NaOH required = 21.30 mL
1) The normality of the acid solution is found as follows;
The chemical reaction is presented as follows;
H₂SO₄(aq) + 2NaOH (aq) → Na₂SO₄ (aq) + H₂O
Number of moles of NaOH in the reaction is found as follows;
[tex]n = \dfrac{21.30}{1,000} \times 0.3200 \, M = \mathbf{0.006816 \, M}[/tex]
Therefore;
The number of moles of H₂SO₄ = 0.006816 M ÷ 2 = 0.003408 M
[tex]Normality = \mathbf{ \dfrac{Mass \ of \, Acid \ in \ reaction}{Equivalent \ mass \times Volume \ of \ soltute}}[/tex]
Which gives;
[tex]Normality = \dfrac{ 98 \times 0.003408 }{49 \times 0.015} = \mathbf{0.4544}[/tex]
The normality of the acid solution, H₂SO₄(aq), N ≈ 0.45442) The molarity is found as follows;
[tex]Molarity = \dfrac{0.003408 \, moles}{0.015 \, L} = \mathbf{0.2272 \, M}[/tex]
The molarity of the acid solution is 0.2272 MLearn more about the normality and the molarity of a solution here:
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A long corridor has a single light bulb and two doors with light switch at each door. design logic circuit for the light; assume that the light is off when both switches are in the same position.
Answer and Explanation:
Let A denote its switch first after that we will assume B which denotes the next switch and then we will assume C stand for both the bulb. we assume 0 mean turn off while 1 mean turn on, too. The light is off, as both switches are in the same place. This may be illustrated with the below table of truth:
A B C (output)
0 0 0
0 1 1
1 0 1
1 1 0
The logic circuit is shown below
C = A'B + AB'
If the switches are in multiple places the bulb outcome will be on on the other hand if another switches are all in the same place, the result of the bulb will be off. This gate is XOR. The gate is shown in the diagram adjoining below.
Commutation is the process of converting the ac voltages and currents in the rotor of a dc machine to dc voltages and currents at its terminals. True False
Answer:
false
Explanation:
the changing of a prisoner sentence or another penalty to another less severe
For a bolted assembly with six bolts, the stiffness of each bolt is kb=Mlbf/in and the stiffness of the members is km=12Mlbf/in. An external load of 80 kips is applied to the entire joint. Assume the load is equally distributed to all the bolts. It has been determined to use 1/2 in- 13 UNC grade 8 bolts with rolled threads. Assume the bolts are preloaded to 75% of the proof load. Clearly state any assumptions.
(a) Determine the yielding factor of safety,
(b) Determine the overload factor of safety,
(c) Determine the factor of safety baserd on joint seperation.
Answer:
nP ≈ 4.9 nL = 1.50Explanation:
GIVEN DATA
external load applied (p) = 85 kips
bolt stiffness ( Kb ) = 3(10^6) Ibf / in
Member stiffness (Km) = 12(10^6) Ibf / in
Diameter of bolts ( d ) = 1/2 in - 13 UNC grade 8
Number of bolts = 6
assumptions
for unified screw threads UNC and UNF
tensile stress area ( A ) = 0.1419 in^2
SAE specifications for steel bolts for grade 8
we have
Minimum proff strength ( Sp) = 120 kpsi
Minimum tensile strength (St) = 150 Kpsi
Load Bolt (p) = external load / number of bolts = 85 / 6 = 14.17 kips
Given the following values
Fi = 75%* Sp*At = (0.75*120*0.1419 ) = 12.771 kip
Preload stress
αi = 0.75Sp = 0.75 * 120 = 90 kpsi
stiffness constant
C = [tex]\frac{Kb}{Kb + Km}[/tex] = [tex]\frac{3}{3+2}[/tex] = 0.2
A) yielding factor of safety
nP = [tex]\frac{sPAt}{Cp + Fi}[/tex] = [tex]\frac{120* 0.1419}{0.2*14.17 + 12.771}[/tex]
nP = 77.028 / 15.605 = 4.94 ≈ 4.9
B) Determine the overload factor safety
[tex]nL = \frac{SpAt - Fi}{CP}[/tex] = ( 120 * 0.1419) - 12.771 / 0.2 * 14.17
= 17.028 - 12.771 / 2.834
= 1.50
Consider a double-pipe counter-flow heat exchanger. In order to enhance its heat transfer, the length of the heat exchanger is doubled. Will the effectiveness of the exchanger double?
Answer:
effectiveness of the heat exchanger will not be double when the length of the heat exchanger is doubled.
Because effectiveness depends on NTU and not necessarily the length of the heat exchanger
For a fluid flowing through a pipe assuming that pressure drop per unit length of pipe (P/L) depends on the diameter of the pipe , the velocity of fluid, the density of fluid and the viscosity of the fluid. Show that = ∅ ൬ ൰
Answer:
Explanation:
La vaca
El pato
As steam is slowly injected into a turbine, the angular acceleration of the rotor is observed to increase linearly with the time t. Know that the rotor starts from rest at t = 0 and that after 10 s the rotor has completed 20 revolutions.Choose the correct equations of motion for the rotor. (You must provide an answer before moving on to the next part.)
a) a = 2kt, w = 3krº, and 8 = 4kr
b) a = {kt, w = ke?, and 0 = }ke?
c) a = kr?, w = jke', and 0 = krº
d) a = kt, w = jke?, and 0 kr
Answer:
α = kt
ω = [tex]\frac{kt^2}{2}[/tex]
θ = [tex]\frac{kt^3}{6}[/tex]
Explanation:
given data
Initial velocity ω = 0
time t = 10 s
Number of revolutions = 20
solution
we will take here first α = kt .....................1
and as α = [tex]\frac{d\omega}{dt}[/tex]
so that
[tex]\frac{d\omega}{dt}[/tex] = kt ..................2
now we will integrate it then we will get
∫dω = [tex]\int_{0}^{t} kt\ dt[/tex] .......................3
so
ω = [tex]\frac{kt^2}{2}[/tex]
and
ω = [tex]\frac{d\theta}{dt}[/tex] ..............4
so that
[tex]\frac{d\theta}{dt}[/tex] = [tex]\frac{kt^2}{2}[/tex]
now we will integrate it then we will get
∫dθ = [tex]\int_{0}^{t}\frac{kt^2}{2} \ dt[/tex] ...............5
solve it and we get
θ = [tex]\frac{kt^3}{6}[/tex]
10 kg/s Propane at 10 bar and 20 C is directed to an adiabatic rigid mixer and is mixed with 20 kg/s Propane at 10 bar and 40 C. What is the final volumetric flow rate in (m3/s) of the resulting mixture.
Answer:
The final volumetric flow rate will be "76.4 m³/s".
Explanation:
The given values are:
[tex]\dot{m_{1}}=10 \ kg/s[/tex]
[tex]\dot{m_{2}}=20 \ Kg/s[/tex]
[tex]T_{1}=293 \ K[/tex]
[tex]T_{2}=313 \ K[/tex]
[tex]P_{1}=P_{2}=P_{3}=10 \ bar[/tex]
As we know,
⇒ [tex]E_{in}=E_{out}[/tex]
[tex]\dot{m_{1}}h_{1}+\dot{m_{2}}h_{2}=\dot{m_{3}}h_{3}[/tex]
[tex]e_{1}\dot{v_{1}}h_{1}+e_{2}\dot{v_{2}}h_{2}=e_{3}\dot{v_{3}}h_{3}[/tex]
[tex]\frac{P_{1}}{RP_{1}}\dot{v_{1}} \ C_{p}T_{1}+ \frac{P_{2}}{RP_{2}}\dot{v_{2}} \ C_{p}T_{1}=\frac{P_{3}}{RP_{3}}\dot{v_{3}} \ C_{p}T_{3}[/tex]
⇒ [tex]\dot{v_{3}}=\dot{v_{1}}+\dot{v_{2}}[/tex]
[tex]=\frac{\dot{m_{1}}}{e_{1}}+\frac{\dot{m_{2}}}{e_{2}}[/tex]
On substituting the values, we get
[tex]=\frac{10}{10\times 10^5}\times 8314\times 293+\frac{20\times 8314\times 313}{10\times 10^5}[/tex]
[tex]=76.4 \ m^3/s[/tex]
why is the peak value of the rectified output less than the peak value of the ac input and by how much g
Answer:
The Peak value of the output voltage is less or lower than that of the peak value of the input voltage by 0.6V reason been that the voltage is tend to drop across the diode.
Explanation:
This is what we called HALF WAVE RECTIFIER in which the Peak value of the output voltage is less or lower than that of the peak value of the input voltage by 0.6V reason been that the voltage is tend to drop across the diode.
Therefore this is the formula for Half wave rectifier
Vrms = Vm/2 and Vdc
= Vm/π:
Where,
Vrms = rms value of input
Vdc = Average value of input
Vm = peak value of output
Hence, half wave rectifier is a rectifier which allows one half-cycle of an AC voltage waveform to pass which inturn block the other half-cycle which is why this type of rectifiers are often been used to help convert AC voltage to a DC voltage, because they only require a single diode to inorder to construct.
The effectiveness of a heat exchanger is defined as the ratio of the maximum possible heat transfer rate to the actual heat transfer rate.
a. True
b. False
Answer:
False
Explanation:
Because
The effectiveness (ϵ) of a heat exchanger is defined as the ratio of the actual heat transfer to the maximum possible heat transfer.
A ramp from an expressway with a design speed of 30 mi/h connects with a local road, forming a T intersection. An additional lane is provided on the local road to allow vehicles from the ramp to turn right onto the local road without stopping. The turning roadway has stabilized shoulders on both sides and will provide for a onelane, one-way operation with no provision for passing a stalled vehicle. Determine the width of the turning roadway if the design vehicle is a single-unit truck. Use 0.08 for superelevation.
Answer:
the width of the turning roadway = 15 ft
Explanation:
Given that:
A ramp from an expressway with a design speed(u) = 30 mi/h connects with a local road
Using 0.08 for superelevation(e)
The minimum radius of the curve on the road can be determined by using the expression:
[tex]R = \dfrac{u^2}{15(e+f_s)}[/tex]
where;
R= radius
[tex]f_s[/tex] = coefficient of friction
From the tables of coefficient of friction for a design speed at 30 mi/h ;
[tex]f_s[/tex] = 0.20
So;
[tex]R = \dfrac{30^2}{15(0.08+0.20)}[/tex]
[tex]R = \dfrac{900}{15(0.28)}[/tex]
[tex]R = \dfrac{900}{4.2}[/tex]
R = 214.29 ft
R ≅ 215 ft
However; given that :
The turning roadway has stabilized shoulders on both sides and will provide for a onelane, one-way operation with no provision for passing a stalled vehicle.
From the tables of "Design widths of pavement for turning roads"
For a One-way operation with no provision for passing a stalled vehicle; this criteria falls under Case 1 operation
Similarly; we are told that the design vehicle is a single-unit truck; so therefore , it falls under traffic condition B.
As such in Case 1 operation that falls under traffic condition B in accordance with the Design widths of pavement for turning roads;
If the radius = 215 ft; the value for the width of the turning roadway for this conditions = 15ft
Hence; the width of the turning roadway = 15 ft
Identify the advantages of using 6 tube passes instead of just 2 of the same diameter in shell-and-tube heat exchanger.What are the advantages and disadvantages of using 6 tube passes instead of just 2 of the same diameter?
Answer:
Please check explanation for answer
Explanation:
Here, we are concerned with stating the advantages and disadvantages of using a 6 tube passes instead of a 2 tube passes of the same diameter:
Advantages
* By using a 6 tube passes diameter, we are increasing the surface area of the heat transfer surface
* As a result of increasing the heat transfer surface area, the rate of heat transfer automatically increases too
Thus, from the above, we can conclude that the heat transfer rate of a 6 tube passes is higher than that of a 2 tube passes of the same diameter.
Disadvantages
* They are larger in size and in weight when compared to a 2 tube passes of the same diameter and therefore does not find use in applications where space conservation is quite necessary.
* They are more expensive than the 2 tube passes of the same diameter and thus are primarily undesirable in terms of manufacturing costs
One kg of an idea gas is contained in one side of a well-insulated vessel at 800 kPa. The other side of the vessel is under vacuum. The two sides are separated by a piston that is initially held in place by the pins. The pins are removed and the gas suddenly expands until it hits the stops. What happens to the internal energy of the gas?
a. internal energy goes up
b. internal energy goes down
c. internal energy stays the same
d. we need to know the volumes to make the calculation
Answer:
Option C = internal energy stays the same.
Explanation:
The internal energy will remain the same or unchanged because this question has to do with a concept in physics or classical chemistry (in thermodynamics) known as Free expansion.
So, the internal energy will be equals to the multiplication of the change in temperature, the heat capacity (keeping volume constant) and the number of moles. And in free expansion the internal energy is ZERO/UNCHANGED.
Where, the internal energy, ∆U = 0 =quantity of heat, q - work,w.
The amount of heat,q = Work,w.
In the concept of free expansion the only thing that changes is the volume.
Compute the volume percent of graphite, VGr, in a 3.2 wt% C cast iron, assuming that all the carbon exists as the graphite phase. Assume densities of 7.9 and 2.3 g/cm3 for ferrite and graphite, respectively.
Answer:
The volume percentage of graphite is 10.197 per cent.
Explanation:
The volume percent of graphite is the ratio of the volume occupied by the graphite phase to the volume occupied by the graphite and ferrite phases. The weight percent in the cast iron is 3.2 wt% (graphite) and 96.8 wt% (ferrite). The volume percentage of graphite is:
[tex]\%V_{gr} = \frac{V_{gr}}{V_{gr}+V_{fe}} \times 100\,\%[/tex]
Where:
[tex]V_{gr}[/tex] - Volume occupied by the graphite phase, measured in cubic centimeters.
[tex]V_{fe}[/tex] - Volume occupied by the graphite phase, measured in cubic centimeters.
The expression is expanded by using the definition of density and subsequently simplified:
[tex]\%V_{gr} = \frac{\frac{m_{gr}}{\rho_{gr}} }{\frac{m_{gr}}{\rho_{gr}}+\frac{m_{fe}}{\rho_{fe}}}\times 100\,\%[/tex]
Where:
[tex]m_{fe}[/tex], [tex]m_{gr}[/tex] - Masses of the ferrite and graphite phases, measured in grams.
[tex]\rho_{fe}, \rho_{gr}[/tex] - Densities of the ferrite and graphite phases, measured in grams per cubic centimeter.
[tex]\%V_{gr} = \frac{1}{1+\frac{\frac{m_{fe}}{\rho_{fe}} }{\frac{m_{gr}}{\rho_{gr}} } }\times 100\,\%[/tex]
[tex]\%V_{gr} = \frac{1}{1 + \left(\frac{\rho_{gr}}{\rho_{fe}} \right)\cdot\left(\frac{m_{fe}}{m_{gr}} \right)} \times 100\,\%[/tex]
If [tex]\rho_{gr} = 2.3\,\frac{g}{cm^{3}}[/tex], [tex]\rho_{fe} = 7.9\,\frac{g}{cm^{3}}[/tex], [tex]m_{gr} = 3.2\,g[/tex] and [tex]m_{fe} = 96.8\,g[/tex], the volume percentage of graphite is:
[tex]\%V_{gr} = \frac{1}{1+\left(\frac{2.3\,\frac{g}{cm^{3}} }{7.9\,\frac{g}{cm^{3}} } \right)\cdot \left(\frac{96.8\,g}{3.2\,g} \right)} \times 100\,\%[/tex]
[tex]\%V_{gr} = 10.197\,\%V[/tex]
The volume percentage of graphite is 10.197 per cent.
Following are the solution to the given points:
[tex]\to C_{Gr} = 100\\\\ \to C_{\alpha}= 0[/tex]From [tex]Fe-F_{\frac{e}{3}} c[/tex] diagram.
[tex]\to W_{\alpha} =\frac{C_{Gr}-C_{o}}{C_{Gr}-C_{\alpha}}[/tex]
[tex]= \frac{100-3.6}{100-0} \\\\= \frac{100-3.6}{100} \\\\= \frac{96.4}{100} \\\\=0.964[/tex]
Calculating the weight fraction of graphite:
[tex]\to W_{Gr}=\frac{C_0 - c_d}{C_{Gr} -c_d}[/tex]
[tex]= \frac{3.6-0}{100-0} \\\\ = \frac{3.6}{100} \\\\= 0.036[/tex]
Calculating the volume percent of graphite:
[tex]\to V_{Gr}=\frac{\frac{W_{Gr}}{P_{Gr}}}{\frac{w_{\alpha}}{P_{\alpha}}+ \frac{W_{Gr}}{P_{Gr}}}[/tex]
[tex]=\frac{\frac{0.036}{2.3}}{\frac{0.964}{7.9}+\frac{0.036}{2.3}}\\\\=0.11368 \times 100\%\\\\=11.368\%[/tex]
Therefore, the final answer is "0.964, 0.036, and 11.368%"
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The value of an SMT capacitor is signified by a
Answer:
Working volttage
Explanation:
SMT electrolytic capacitors are marked with working voltage. The value of these capacitors is measured in micro farads. It is a surface mount capacitor which is used for high volume manufacturers. They are small lead less and are widely used. They are placed on modern circuit boards.
Air enters the first compressor stage of a cold-air standard Brayton cycle with regeneration and intercooling at 100 kPa, 300 K, with a mass flow rate of 6 kg/s. The overall compressor pressure ratio is 10, and the pressure ratios are the same across each compressor stage. The temperature at the inlet to the second compressor stage is 300 K. The turbine inlet temperature is 1400 K. The compressor stages and turbine each have isentropic efficiencies of 80% and the regenerator effectiveness is 80%. For k = 1.4, calculate:
a. the thermal efficiency of the cycle
b. the back work ratio
c. the net power developed, in kW
d. the rates of exergy destruction in each compressor stage and the turbine stage as well as the regenerator, in kW, for T 0 = 300 K.
Answer:
a. [tex]\eta _{th}[/tex] = 77.65%
b. bwr = 6.5%
c. 3538.986 kW
d. -163.169 kJ
Explanation:
a. The given property are;
P₂/P₁ = 10, P₂ = 10 * 100 kPa = 1000 kPa
p₄/p₁ = 10
P₂/P₁ = p₄/p₃ = √10
p₂ = 100·√10
[tex]T_{2s}[/tex] = T₁×(√10)^(0.4/1.4) = 300 × (√10)^(0.4/1.4) = 416.85 K
T₂ = T₁ + ([tex]T_{2s}[/tex] - T₁)/[tex]\eta _c[/tex] = 300 + (416.85 - 300)/0.8 = 446.0625 K
p₄ = 10×p₁ = 10×100 = 1000 kPa
p₄/p₃ = √10 =
p₃ = 100·√10
T₃ = 300 K
T₃/[tex]T_{4s}[/tex] = (P₂/P₁)^((k - 1)/k) = (√10)^(0.4/1.4)
[tex]T_{4s}[/tex] = T₃/((√10)^(0.4/1.4) ) = 300/((√10)^(0.4/1.4)) = 215.905 K
T₄ = T₃ + ([tex]T_{4s}[/tex] - T₃)/[tex]\eta _c[/tex] = 300 + (215.905- 300)/0.8 = 194.881 K
The efficiency = 1 - (T₄ - T₁)/(T₃ - T₂) = 1 - (194.881 -300)/(300 -446.0625 ) = 0.28
T₄ = 446.0625 K
T₆ = 1400 K
[tex]T_{7s}[/tex]/T₆ = (1/√10)^(0.4/1.4)
[tex]T_{7s}[/tex] = 1400×(1/√10)^(0.4/1.4) = 1007.6 K
T₇ = T₆ - [tex]\eta _t[/tex](T₆ - [tex]T_{7s}[/tex]) = 1400 - 0.8*(1400 - 1007.6) = 1086.08 K
T₈ = 1400 K
T₉ = 1086.08 K
T₅ = T₄ + [tex]\epsilon _{regen}[/tex](T₉ - T₄) = 446.0625 +0.8*(1086.08 - 446.0625) = 958.0765 K
[tex]\eta _{th}[/tex] =(((T₆ - T₇) + (T₈ - T₉)) -((T₂ - T₁) + (T₄ - T₃)))/((T₆ - T₅) + (T₈ - T₇))
(((1400 - 1086.08) + (1400 -1086.08 ))-((446.0625 - 300)+(194.881 - 300)))/((1400 -958.0765 ) + (1400 -1086.08 )) = 0.7765
[tex]\eta _{th}[/tex] = 77.65%
b. Back work ratio, bwr = [tex]bwr = \dfrac{w_{c,in}}{w_{t,out}}[/tex]
((446.0625 - 300)+(194.881 - 300))/((1400 - 1086.08) + (1400 -1086.08 ))
40.9435/627.84 = 6.5%
c. [tex]w_{net, out} = c_p[(T_6 -T_7) + (T_8 - T_9)] - [(T_2 - T_1) + (T_4 -T_3)][/tex]
Power developed is given by the relation;
[tex]\dot m \cdot w_{net, out}[/tex]
[tex]\dot m \cdot w_{net, out}[/tex]= 6*1.005*(((1400 - 1086.08) + (1400 -1086.08 ))-((446.0625 - 300)+(194.881 - 300))) = 3538.986 kW
d. Exergy destruction = 6*(1.005*(300-446.0625 ) - 300*1.005*(-0.3966766)
-163.169 kJ
. The job of applications engineer for which Maria was applying requires (a) excellent technical skills with respect to mechanical engineering, (b) a commitment to working in the area of pollution control, (c) the ability to deal well and confidently with customers who have engineering problems, (d) a willingness to travel worldwide, and (e) a very intelligent and well-balanced personality. List 10 questions you would ask when interviewing applicants for the job.
Answer:
Tell us about your self Are your confident that you are the right candidate for this positionwhy should i hire youDo you like working under supervisionHow do you like to work ( in a group or individually )What is your ultimate workplace goalwhat are your future plansWhat do you expect from the Organization when given the jobDo you like taking on critical problemsHow long can you work in this positionExplanation:
For a job of applications engineer which require excellent technical skills, commitment to working , ability to deal well and confidently with customers a willingness to travel and very intelligent and well-balanced personality.
The ten questions you should ask Maria to determine if she is qualified for the job are :
Tell us about your self ( functions you have )Are your confident that you are the right candidate for this positionwhy should i hire youDo you like working under supervisionHow do you like to work ( in a group or individually )What is your ultimate workplace goalwhat are your future plansWhat do you expect from the Organization when given the jobDo you like taking on critical problemsHow long can you work in this position